Define a function $F: \mathbb R \to \mathbb C$ $$F(x) = \log(-i(x+ia)),$$ where $a=0^+$ is a regulator, and the logarithm is always in the principal branch.
$F$ is a tempered distribution: for any $g\in \mathcal S(\mathbb R)$, $$\langle F,g \rangle = \lim_{a\to0^+} \int_{-\infty}^\infty F(x) g(x) \: dx$$ is well-defined and $g\mapsto \langle F,g \rangle$ is continuous. (To prove this, we adopt the method in log|x| is a tempered distribution)
Question: What is the Fourier transform $$\hat F(k) = \int_{-\infty}^\infty F(x) e^{-ikx} dx$$ of $F$?
Remark: Differentiation gives $F'(x) = \frac{1}{x+ia}$. This is a well-known distribution whose Fourier transform is $\widehat{F'}(k) = -2\pi i \Theta(k)$, where $\Theta$ is the Heaviside function. Hence, using the property $\widehat{F'}(k) = ik\widehat F(k)$, we have $$\widehat F(k) = -\frac{2\pi}{k} \Theta(k) + C\delta(k),$$ where $C$ is a constant. Hence, the problem reduces to determine $C$.